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I'm repeating a bit of set theory these days and was a bit perplexed by the axiom schema of replacement, for which I can't really seem to find a substantial application. After all, if $F: A \to B$ is a function between sets, the image will exist due to separation.

I read on WP about some implications, but they all seem of limited practical interest for day-to-day maths. So why do we usually include this axiom?

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    $\begingroup$ Do you view the existence of ordinals $\geq \omega+\omega$ to be of limited practical interest for day-to-day maths? $\endgroup$
    – Alex Kruckman
    Mar 29 at 15:32
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    $\begingroup$ I think it depends what you mean by "day-to-day" math. $V_{\omega+\omega}$ satisfies all of ZFC except for replacement, and a huge amount of the math people do can be carried out there. In that sense, replacement is unnecessary. On the other hand, having replacement around is convenient. Suppose you want to attach a ring to each open set in a topological space $(X, \tau)$. You give a map $F : \tau \to \mathbf{Rings}$. In any reasonable case you could go through the effort of finding a set-sized object to replace $\mathbf{Rings}$, but with replacement that effort is deemed unnecessary. $\endgroup$
    – Chris Eagle
    Mar 29 at 15:52
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    $\begingroup$ The point is that without replacement you can't generally recurse along an arbitrary well-order, which limits the use of well-orders in the first place. That said, it is true that the vast majority of mathematics goes through in ZC (= set theory without replacement) or indeed much less. This MO discussion may be of interest. $\endgroup$
    – Noah Schweber
    Mar 29 at 15:52
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    $\begingroup$ The construction you describe gives a well-order that is isomorphic to the ordinal $\omega+\omega$, but it's not the ordinal itself. The question of why we might actually want to work with the von Neumann ordinals instead of building sufficiently large well-orders in ad hoc ways was asked recently on StackExchange and MathOverflow. You might find the answers there interesting. I'm obligated to also link to Kanamori's wonderful article In Praise of Replacement. $\endgroup$
    – Alex Kruckman
    Mar 29 at 15:54
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    $\begingroup$ In addition to the links given by @Alex, I wrote karagila.org/2019/in-praise-of-replacement a few years ago. $\endgroup$
    – Asaf Karagila ♦
    Mar 30 at 7:31

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Since all the answers went straight into the comment section, let me here summarise the arguments that were made in favour of the axiom schema of replacement. I want to begin with one that I just came up with. All the other items are taken from the comment section.

  • It is not unreasonable that the "images" of "class functions" are also sets, because everything which is not a set is supposed to be "larger" than a set, but these "images" are "equal or smaller".
  • Certain constructions (such as associating an element of $\operatorname{Ring}$ to each point $x$ of a topological space $X$) are easier if the axiom schema is being regarded as true
  • The axiom schema allows for the construction of von Neumann ordinals, which we want because they exist in any universe set class
  • Transfinite recursion is impossible without the axiom schema

I have to admit that the second item is the one that convinces me the most, and the others are more like "added bonuses". Constructing a suitable set of rings in such a situation seems unnecessarily inconvenient to me.

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    $\begingroup$ "Transfinite recursion is impossible without the axiom schema" is quite an overstatement. Recursion in full generality (transfinite or even just ordinary recursion on $\mathbb{N}$) requires Replacement. But the vast majority of ordinary applications of transfinite recursion do not need Replacement. The point is that you need Replacement to recursively define a function if you do not already know that the codomain of the putative function is a set. $\endgroup$
    – Eric Wofsey
    Mar 30 at 2:10
  • $\begingroup$ Perhaps one might change it into "transfinite recursion may become more cumbersome" without the axiom schema. $\endgroup$
    – Cloudscape
    Mar 30 at 12:06
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I've thought of it as: the other axioms are useful for "day-to-day math", but Replacement and Foundation are added because they make set theory more beautiful. Since the set theorists are the ones who came up with the axioms, they've earned a reward :)

Indeed, set theory without Replacement is ugly. It's interesting to see what happens in the set $R(\omega + \omega)$, the set of all sets of rank less than $\omega + \omega$, where all of ZC holds but Replacement fails. We have enough sets to do all of everyday arithmetic, analysis, etc, but we have the following facts, all of which are therefore consistent with the absence of Replacement:

  • Every set admits a well ordering, but many of them (e.g. $\mathbb{R}$) are not bijective to any ordinal (since all ordinals are countable). So you lose the canonical representative of a well-ordering type.

  • As a result, you also lose the canonical representative of a cardinality. You can't define $|A|$ as "the least ordinal in bijection with $A$" because there might not be any. So it's no longer clear what objects in your universe to use as your cardinal numbers.

  • Hartogs' lemma is false: every ordinal injects into $\omega$.

  • There exists exactly one limit ordinal, namely $\omega$. No transfinite hierarchy for you.

  • Ordinal arithmetic no longer works, because you cannot add $\omega$ to $\omega$, let alone multiply.

  • For analysis, you have $\mathbb{R}$, you have subsets and functions on $\mathbb{R}$, you have sets of sets and sets of functions, you have operators on function spaces, etc. But you can't take all those objects and put them into one single set; they are a proper class.

  • Even analysis can get a little tricky. Let's take tensor products of Hilbert spaces, so that $H \otimes K$ is a vector space with a bilinear map from $H \times K$, and define higher tensor products recursively as $H \otimes K \otimes L = (H \otimes K) \otimes L$. Then, unfortunately, Fock space doesn't exist, so quantum mechanics becomes awkward. (The problem was that because of the recursion, the set $H^{\otimes k}$ has a rank that increases with $k$, and so we can't put all of them in a set. We could fix it with a non-recursive definition, but I don't think most analysts want to have to take that kind of care.)

  • Goodstein's theorem is still true, but the proof no longer works.

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    $\begingroup$ Several of these points are pretty artificial, though--they can easily be circumvented by just using appropriate well-ordered sets instead of ordinals. For instance, Hartogs' lemma is true if it is stated in terms of arbitrary well-ordered sets rather than ordinals. $\endgroup$
    – Eric Wofsey
    Mar 30 at 2:26
  • $\begingroup$ @EricWofsey: Is there a natural "backup" choice for canonical representatives, then? If I want a set to represent the order type $\omega + \omega$, or $\omega_1$, what should I use? Or uncountable cardinalities? $\endgroup$
    – Nate Eldredge
    Mar 30 at 2:29
  • $\begingroup$ There isn't any particularly canonical choice, but if you have a cardinality bound on your well-orderings, you can just use isomorphism classes of well-orderings restricted to some fixed domain. For instance, you can define $\omega_1$ as the set of isomorphism classes of well-orderings of subsets of $\mathbb{N}$. $\endgroup$
    – Eric Wofsey
    Mar 30 at 2:31
  • $\begingroup$ Thanks for the further input. I now think that in total, especially given Chris Eagle's remark, the benefits of replacement outweigh the cost of having an additional axiom. In the end, brevity of the total exposition is a major criterion. $\endgroup$
    – Cloudscape
    Mar 30 at 12:11

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